0.79/0.85 YES 0.79/0.85 0.79/0.85 DP problem for innermost termination. 0.79/0.85 P = 0.79/0.85 f5#(x1, x2) -> f4#(x1, x2) 0.79/0.85 f4#(I0, I1) -> f3#(0, rnd2) [rnd2 = rnd2] 0.79/0.85 f3#(I2, I3) -> f1#(I2, I3) 0.79/0.85 f1#(I4, I5) -> f3#(1 + I4, I5) [1 + I4 <= 10] 0.79/0.85 R = 0.79/0.85 f5(x1, x2) -> f4(x1, x2) 0.79/0.85 f4(I0, I1) -> f3(0, rnd2) [rnd2 = rnd2] 0.79/0.85 f3(I2, I3) -> f1(I2, I3) 0.79/0.85 f1(I4, I5) -> f3(1 + I4, I5) [1 + I4 <= 10] 0.79/0.85 f1(I6, I7) -> f2(I6, I7) [10 <= I6] 0.79/0.85 0.79/0.85 The dependency graph for this problem is: 0.79/0.85 0 -> 1 0.79/0.85 1 -> 2 0.79/0.85 2 -> 3 0.79/0.85 3 -> 2 0.79/0.85 Where: 0.79/0.85 0) f5#(x1, x2) -> f4#(x1, x2) 0.79/0.85 1) f4#(I0, I1) -> f3#(0, rnd2) [rnd2 = rnd2] 0.79/0.85 2) f3#(I2, I3) -> f1#(I2, I3) 0.79/0.85 3) f1#(I4, I5) -> f3#(1 + I4, I5) [1 + I4 <= 10] 0.79/0.85 0.79/0.85 We have the following SCCs. 0.79/0.85 { 2, 3 } 0.79/0.85 0.79/0.85 DP problem for innermost termination. 0.79/0.85 P = 0.79/0.85 f3#(I2, I3) -> f1#(I2, I3) 0.79/0.85 f1#(I4, I5) -> f3#(1 + I4, I5) [1 + I4 <= 10] 0.79/0.85 R = 0.79/0.85 f5(x1, x2) -> f4(x1, x2) 0.79/0.85 f4(I0, I1) -> f3(0, rnd2) [rnd2 = rnd2] 0.79/0.85 f3(I2, I3) -> f1(I2, I3) 0.79/0.85 f1(I4, I5) -> f3(1 + I4, I5) [1 + I4 <= 10] 0.79/0.85 f1(I6, I7) -> f2(I6, I7) [10 <= I6] 0.79/0.85 0.79/0.85 We use the reverse value criterion with the projection function NU: 0.79/0.85 NU[f1#(z1,z2)] = 10 + -1 * (1 + z1) 0.79/0.85 NU[f3#(z1,z2)] = 10 + -1 * (1 + z1) 0.79/0.85 0.79/0.85 This gives the following inequalities: 0.79/0.85 ==> 10 + -1 * (1 + I2) >= 10 + -1 * (1 + I2) 0.79/0.85 1 + I4 <= 10 ==> 10 + -1 * (1 + I4) > 10 + -1 * (1 + (1 + I4)) with 10 + -1 * (1 + I4) >= 0 0.79/0.85 0.79/0.85 We remove all the strictly oriented dependency pairs. 0.79/0.85 0.79/0.85 DP problem for innermost termination. 0.79/0.85 P = 0.79/0.85 f3#(I2, I3) -> f1#(I2, I3) 0.79/0.85 R = 0.79/0.85 f5(x1, x2) -> f4(x1, x2) 0.79/0.85 f4(I0, I1) -> f3(0, rnd2) [rnd2 = rnd2] 0.79/0.85 f3(I2, I3) -> f1(I2, I3) 0.79/0.85 f1(I4, I5) -> f3(1 + I4, I5) [1 + I4 <= 10] 0.79/0.85 f1(I6, I7) -> f2(I6, I7) [10 <= I6] 0.79/0.85 0.79/0.85 The dependency graph for this problem is: 0.79/0.85 2 -> 0.79/0.85 Where: 0.79/0.85 2) f3#(I2, I3) -> f1#(I2, I3) 0.79/0.85 0.79/0.85 We have the following SCCs. 0.79/0.85 0.79/3.83 EOF